Answer
$(1,1,-7)$
Work Step by Step
Given Equations are
$x+3y+z=-3$ Equation $(1)$
$-x+y+2z=-14$ Equation $(2)$
$3x+2y-z=12$ Equation $(3)$
Adding Equation $(1)$ and Equation $(2)$
$x+3y+z-x+y+2z=-3-14$
$4y+3z=-17$ Equation $(4)$
Adding Equation $(1)$ and Equation $(3)$
$x+3y+z+3x+2y-z=-3+12$
$4x+5y=9$ Equation $(5)$
Multiplying Equation $(2)$ by $3$ then adding with Equation $(3)$
$3(-x+y+2z)+(3x+2y-z)=3(-14)+12$
$-3x+3y+6z+3x+2y-z=-42+12$
$5y+5z=-30$
$y+z=-6$ Equation $(6)$
Multiplying Equation $(6)$ by $-3$ then adding with Equation $(4)$
$-3(y+z)+4y+3z=-3(-6)-17$
$-3y-3z+4y+3z=18-17$
$y=1$
Substituting $y$ value in Equation $(6)$ we get,
$y+z=-6$
$1+z=-6$
$z=-6-1$
$z=-7$
Substituting $y$ value in Equation $(5)$ we get,
$4x+5y=9$
$4x+5(1)=9$
$4x=9-5$
$4x=4$
$x=1$
Solution $(1,1,-7)$
Substituting $x,y,z$ values in
$\frac{x}{2} + \frac{y}{3} + \frac{z}{9} = \frac{1}{18}$
$\frac{1}{2} + \frac{1}{3} - \frac{7}{9} = \frac{1}{18}$
$\frac{9+6-14}{18} = \frac{1}{18}$
$\frac{15-14}{18} = \frac{1}{18}$
$\frac{1}{18} = \frac{1}{18}$